By what Denis said we get that $lim_{x\to\infty}f(x)=\infty$ And so $lim_{x\to\infty}f(f(x))=\infty$.
Now we will reach a contradiction by using the definition of limits at infinity:
Because $lim_{x\to\infty}f(x)=\infty$ and $lim_{x\to\infty}f(f(x))=\infty$. We get that:
$\forall 0<M_1,\exists 0<N_1$ such that $\forall N_1<x, M_1<f(x)$ and that
$\forall 0<M_2,\exists 0<N_2$ such that $\forall N_2<x, M_2<f(f(x))$
Now take $M_1=M_2=1$ and we will get that $\exists 0<N_1$ such that $\forall N_1<x, 1<f(x)$ and that $\exists 0<N_2$ such that $\forall N_2<x, 1<f(f(x))$
Take $x=max (N_1+1,N_2+1)$, then $N_1,N _2<x$ and so we get that $1<f(x)$ and that $1<f(f(x))$
Now take $y=max(\frac{x+1}{f( f( x))-1},N_2-x)$ which implies that $N_2< x+y+1\leq y f(f(x ))$ and so $1<f(f(x+y+1))$, Now $f (x+y)\geq f(x )+yf(f(x))\geq x+y+1$ And hence
$f(f(x+y))\geq f(x+y+1)+(f(x+y)-(x+y+1))f(f(x+y+1)) \geq f(x +y+1)\geq f(x+y)+f(f(x+y))\geq f(x)+yf(f(x))+f(f(x+y))>f(f(x+y))$
And so we get that $0>0$ which is a contradiction.